Problem: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}-8x+5y &= 5 \\ 7x-5y &= -1\end{align*}$
Solution: Begin by moving the $y$ -term in the second equation to the right side of the equation. $7x = 5y-1$ Divide both sides by $7$ to isolate $x$ $x = {\dfrac{5}{7}y - \dfrac{1}{7}}$ Substitute this expression for $x$ in the first equation. $-8({\dfrac{5}{7}y - \dfrac{1}{7}}) + 5y = 5$ $-\dfrac{40}{7}y + \dfrac{8}{7} + 5y = 5$ Simplify by combining terms, then solve for $y$ $-\dfrac{5}{7}y + \dfrac{8}{7} = 5$ $-\dfrac{5}{7}y = \dfrac{27}{7}$ $y = -\dfrac{27}{5}$ Substitute $-\dfrac{27}{5}$ for $y$ in the top equation. $-8x+5( -\dfrac{27}{5}) = 5$ $-8x-27 = 5$ $-8x = 32$ $x = -4$ The solution is $\enspace x = -4, \enspace y = -\dfrac{27}{5}$.